Question

Air of mass 2 kg is compressed reversibly and adiabatically from an initial state at 1.2 kPa, 28°C
to 2 MPa and then expanded at a constant pressure to the original volume.
(i) Sketch the process on the p-V plane;
(ii) Determine the heat and work transfer.
Take for air, assume R = 0.287 kJ/kgK and Cv = 0.713 kJ/kgK

Answer 1

$m=2kg$

$\nu(Air) =28,97 g/mol$

$p_0=1.2kPa$

$P_1=2MPa$

$T_0=28 C$

Solution

$$pV= RT \eta$$

$m=2kg$

$\nu(Air) =28,97 g/mol$

$p_0=1.2kPa$

$P_1=2MPa$

$T_0=28 C$

$pV=\nu RT$ -> $p_0V_0=\nu RT_0$ -> $V_0=\nu RT_0/p_0$

Adiabatic process: $p_0V_0^\gamma=p_1V_1^\gamma$ -> $V_1=...$

where $\gamma=(i+2)/i$ for air $i=1.4$ $\gamma=2.43$

$A_12=\int pdV=p_0\int(V_0/V_1)^\gamma dV= =-p_0V_0^(\gamma-1)* (V_1^(1-\gamma)-V_0^(1-\gamma))$

$A_23=0$

$Q_12=0$

$Q_23=mC_p(T_3-T_2)$

$C_p=C_v+R$

$T_2=p_1V_1/(p_0V_0)T_0$

$T_3=p_1V_0/(p_1V_1)T_2$